Unformatted text preview: k has appropriate dimensions (namely, kg/s 2 ), the force is 5(0 . 1. 08) = 5(0 . 02) = 0 . 1 Newtons. EXAMPLE 9.17 How much work is done in compressing the spring in the previous example from its natural length to 0 . 08 meters? From 0 . 08 meters to 0 . 05 meters? How much work is done to stretch the spring from 0 . 1 meters to 0 . 15 meters? We can approximate the work by dividing the distance that the spring is compressed into small subintervals. Then the force exerted by the spring is approximately constant over the subinterval, so the work required to compress the spring from x i to x i +1 is approximately 5( x i. 1)Δ x . The total work is approximately n1 X i =0 5( x i. 1)Δ x...
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 Fall '07
 JonathanRogawski
 Math, Calculus, Exponential Function, Force, Mass, ........., Mathematical constant

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